Question

1. a sample of 31 boxes of cereal has a sample standard deviation of 0.81 ounces. construct a 95% confidence interval to estimate the true standard deviation of the filling process for the boxes of cereal. a) (0.525, 1.095) b) (0.419, 1.172) c) (0.647, 1.083) d) (0.719, 0.719) e) none of the above.

Explanation:

Step1: Determine the degrees of freedom and critical value

The degrees of freedom \(df=n - 1=31-1 = 30\). For a 95% confidence interval, the significance level \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\). Looking up in the chi - square table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.975,30}^{2}=16.791\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.025,30}^{2}=46.979\)

Step2: Calculate the confidence interval

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{R}^{2}}}\leq\sigma\leq\sqrt{\frac{(n - 1)s^{2}}{\chi_{L}^{2}}}\)

Given \(n = 31\), \(s = 0.81\)

First, calculate \((n - 1)s^{2}=(31 - 1)\times(0.81)^{2}=30\times0.6561 = 19.683\)

Then, \(\sqrt{\frac{19.683}{46.979}}\leq\sigma\leq\sqrt{\frac{19.683}{16.791}}\)

\(\sqrt{0.419}\leq\sigma\leq\sqrt{1.172}\)

\(0.647\leq\sigma\leq1.083\)

Answer:

C) \((0.647,1.083)\)